[Math] Volume of a cuboid whose diagonal and surface area is known

geometryvolume

The sum of length, breadth and depth of cuboid is $19$cm and its diagonal is $5\sqrt{5}$cm.
Its volume is:


a) 125
b) 236
c) 361
d) 486

Solution:

$$\ell^2 + b^2 +h^2 = 125\quad\text{ and }\quad \ell+b+h=19,$$

How can i find volume from this information?

Best Answer

Define $$\begin{cases} p_1 = & \ell + b + h\\ p_2 = & \ell^2 + b^2 + h^2\\ p_3 = & \ell^3 + b^3 + h^3 \end{cases} \quad\text{ and }\quad \begin{cases} s_1 = & \ell + b + h\\ s_2 = & \ell b + \ell h + b h\\ s_3 = & \ell b h \end{cases} $$ We know $p_1 = s_1 = 19$ and $p_2 = 125$.

Apply $AM \ge GM$ to the three numbers $\ell, b, h$, we get an upper bound for $s_3$:

$$s_3 = \ell b h \le \left(\frac{\ell + b + h}{3}\right)^3 = \frac{p_1^3}{3} = \frac{19}{3}^3 \sim 254.037$$ This rules out choices (c) and (d).

Apply Cauchy Schwarz inequality to $( \sqrt{\ell}, \sqrt{b}, \sqrt{h} )$ and $( \sqrt{\ell}^3, \sqrt{b}^3, \sqrt{h}^3 )$, we get

$$p_2^2 \le p_1 p_3$$

Together with the Newton's identities

  • $p_1 = s_1$,
  • $p_2 = s_1 p_1 - 2 s_2$,
  • $p_3 = s_1 p_2 - s_2 p_1 + 3s_3$

We obtain a lower bound for $s_3$:

$$s_3 = \frac13 \left( p_3 - s_1 p_2 + s_2 p_1 \right) \ge \frac13 \left( \frac{p_2^2}{p_1} - p_1 p_2 + \frac{p_1^2 - p_2}{2} p_1 \right) = \frac13 \left( \frac{p_2^2}{p_1} + \frac{p_1^2 - 3 p_2}{2} p_1 \right) = \frac13 \left( \frac{125^2}{19} + \frac{19^2 - 3\cdot 125}{2} \cdot 19\right) = \frac{4366}{19} \sim 229.790 $$ This rules out choice (a) and leaves us choice (b) $s_3 = 236$ as the only possible answer.

Related Question